A first order reaction undergoes 50,% complet | vedclass

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(C) For a first order reaction,the half-life $t_{1/2} = 50\, min$.The rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{50\, min}$.For $80\%$ co

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$117$

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(C) For a first order reaction,the half-life $t_{1/2} = 50\, min$.The rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{50\, min}$.For $80\%$ completion,the remaining concentration is $100\% - 80\% = 20\%$.The time $t$ is given by the formula: $t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} ight)$.Substituting the values: $t = \frac{2.303}{(0.693/50)} \log \left( \frac{100}{20} ight)$.$t = \frac{2.303 	imes 50}{0.693} 	imes \log(5)$.Since $\log(5) \approx 0.699$ and $\frac{2.303}{0.693} \approx 3.32$,we get $t \approx 3.32 	imes 50 	imes 0.699 \approx 116.1\, min$.Rounding to the nearest integer,$t \approx 117\, min$.

$A$ first order reaction undergoes $50\,\%$ completion in $50\, min$,then it undergoes $80\, \%$ completion in ........... $min$

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